Problem: Let $f(x) = 2x^{2}+10x+8$. Where does this function intersect the x-axis (i.e. what are the roots or zeroes of $f(x)$ )?
Answer: The function intersects the x-axis when $f(x) = 0$ , so you need to solve the equation: $2x^{2}+10x+8 = 0$ Use the quadratic formula to solve $ax^2 + bx + c = 0$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $a = 2, b = 10, c = 8$ $ x = \dfrac{-10 \pm \sqrt{10^{2} - 4 \cdot 2 \cdot 8}}{2 \cdot 2}$ $ x = \dfrac{-10 \pm \sqrt{36}}{4}$ $ x = \dfrac{-10 \pm 6}{4}$ $x =-1,-4$